Question:

Given the complex number $z=\frac{1}{2}-2i$ and the equation $|z+1|=\alpha z+\beta(1+i)$, where $\alpha,\beta\in\mathbb{R}$. Step 1: Calculate the value of $|z+1|$. First, find $z+1$: $$z+1 = \left(\frac{1}{2}-2i\right)+1 = \frac{3}{2}-2i$$ Now, calculate its modulus: $$|z+1| = \left|\frac{3}{2}-2i\right| = \sqrt{\left(\frac{3}{2}\right)^2 + (-2)^2} = \sqrt{\frac{9}{4}+4} = \sqrt{\frac{9+16}{4}} = \sqrt{\frac{25}{4}} = \frac{5}{2}$$ Step 2: Express the right side of the equation, $\alpha z+\beta(1+i)$, in the form $A+Bi$. Substitute the value of $z$: $$\alpha z+\beta(1+i) = \alpha\left(\frac{1}{2}-2i\right) + \beta(1+i)$$ $$= \left(\frac{\alpha}{2} - 2\alpha i\right) + (\beta + \beta i)$$ $$= \left(\frac{\alpha}{2} + \beta\right) + (-2\alpha + \beta)i$$ Step 3: Equate the real and imaginary parts. The given equation is $|z+1|=\alpha z+\beta(1+i)$. Substituting the results from Step 1 and Step 2: $$\frac{5}{2} = \left(\frac{\alpha}{2} + \beta\right) + (-2\alpha + \beta)i$$ For a real number to be equal to a complex number, the imaginary part of the complex number must be zero, and its real part must be equal to the real number. This gives us a system of two equations: 1. Imaginary part: $-2\alpha + \beta = 0$ 2. Real part: $\frac{\alpha}{2} + \beta = \frac{5}{2}$ Step 4: Solve for $\alpha$ and $\beta$. From equation (1), we have $\beta = 2\alpha$. Substitute this into equation (2): $$\frac{\alpha}{2} + (2\alpha) = \frac{5}{2}$$ $$\frac{\alpha}{2} + \frac{4\alpha}{2} = \frac{5}{2}$$ $$\frac{5\alpha}{2} = \frac{5}{2}$$ Multiplying both sides by $\frac{2}{5}$ yields: $$\alpha = 1$$ Now, substitute $\alpha=1$ back into $\beta=2\alpha$: $$\beta = 2(1) = 2$$ Step 5: Calculate $\alpha+\beta$. $$\alpha+\beta = 1+2 = 3$$